If a ring has a 1 (i.e. a ring has a multiplicative inverse), does this imply that the subring has a 1 as well?

Thank you

  • $\begingroup$ multiplicative identity? Look at $\Bbb Z$ and $2 \Bbb Z$ $\endgroup$ – Chinnapparaj R Apr 1 at 15:15
  • $\begingroup$ Think about the zero divisors. $\endgroup$ – Μάρκος Καραμέρης Apr 1 at 15:15
  • $\begingroup$ Yes. Subrings of rings with identity contain the multiplicative identity. There are authors that say otherwise so, ultimately, you have to consult your textbook or your teacher. I disagree with those authors because a sub$thing$ must preserve the structure of the $thing$. Since $1$ is part of the structure of a ring with identity it should be preserved by subrings. $\endgroup$ – John Douma Apr 1 at 15:21

If your definition of "subring $S$ of $R$" includes verbiage saying "and it has to have the same identity as $R$" then trivially yes.

But some authors do not require that, and some don't require identities at all. So, it just depends on what definitions you are using.

An abelian subgroup that is closed under multiplication could have no identity element, or it could have an identity distinct from that of $R$, or it could have the same identity as $R$. Which one of these you call "a subring" is up to the definitions you've settled on.


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